Q.What is the de Broglie wavelength associated with
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De Broglie Wavelength: When Particles Start Acting Like Waves
You already know that light behaves like a wave (interference, diffraction) and like a particle (photoelectric effect). That's wave-particle duality for light. De Broglie's radical idea in 1924 was: if light can be both, why can't matter be both too?
He proposed that every moving particle — an electron, a proton, even a cricket ball — has a wavelength associated with it. The faster it moves, the shorter that wavelength becomes.
The Intuition
Think of a wave on a string. Its wavelength is the distance between two consecutive crests. Now imagine an electron moving through space. De Broglie said that the electron's motion itself creates a "matter wave" — a wave of probability that guides where the electron is likely to be found.
You never see this wavelength in everyday life because for large objects it's unimaginably tiny. A cricket ball moving at 30 m/s has a de Broglie wavelength of about 10−34 m — far smaller than an atomic nucleus. That's why macroscopic objects behave like particles.
The Precise Statement
The de Broglie wavelength λ of a particle is given by:
λ=ph
where:
- h is Planck's constant (6.626×10−34 J⋅s)
- p is the momentum of the particle (p=mv for non-relativistic speeds)
Key point: The wavelength depends only on momentum, not on charge, mass, or any other property. A fast electron and a slow proton can have the same wavelength if their momenta are equal.
What This Means Physically
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For electrons in atoms: The de Broglie wavelength of an electron in a hydrogen atom is roughly the size of the atom itself (≈10−10 m). This is why electrons form standing waves around the nucleus — only certain wavelengths "fit" into the orbit, which explains quantised energy levels.
-
For experiments: If you fire electrons through a crystal, they diffract just like X-rays. This was confirmed by Davisson and Germer in 1927 — a Nobel-winning experiment that proved de Broglie right.
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For large objects: The wavelength is so small that wave behaviour is undetectable. A car moving at 100 km/h has λ≈10−38 m — you'd need a slit smaller than an atom to see diffraction.
A common mistake is to think the de Broglie wavelength is the size of the particle. It is not. It is the wavelength of the probability wave associated with the particle. The particle itself remains point-like.
Worked Example
Question: What is the de Broglie wavelength of an electron moving at 2.0×106 m/s? (Mass of electron me=9.11×10−31 kg)
Solution:
First, find momentum:
p=mv=(9.11×10−31)(2.0×106)=1.822×10−24 kg⋅m/s
Then apply de Broglie's formula: …
Why this formula?
De Broglie Wavelength: Why Matter Has a Wave Nature
The idea that a moving particle has a wavelength is one of the most radical shifts in physics. It came from Louis de Broglie in 1924, who asked a simple question: if light — which we thought was a wave — can behave like a particle (the photon), then why can't a particle behave like a wave?
The Core Insight: Symmetry in Nature
De Broglie started from Einstein's relation for a photon. For light, the energy E and momentum p of a photon are linked to its wave properties — frequency f and wavelength λ — by:
E=hfandp=λh
where h is Planck's constant. These are not arbitrary; they come from the fact that light is an electromagnetic wave, and Planck had already shown that energy comes in quanta hf.
De Broglie's reasoning was a leap of symmetry: if nature treats light and matter on equal footing (as Einstein's special relativity suggests), then any moving particle should also have a wavelength associated with it. He proposed that the same relation holds for matter:
λ=ph
where p=mv is the momentum of the particle (for non-relativistic speeds). This is the de Broglie wavelength.
Why This Formula Makes Sense: A Simple Derivation
There is no rigorous "derivation" from first principles — de Broglie's hypothesis was a postulate. But we can see why it is plausible by combining two key ideas from relativity and quantum theory.
Step 1: Energy of a particle from relativity
For a particle with rest mass m0, the total energy in special relativity is:
E=p2c2+m02c4
For a photon, m0=0, so E=pc. This matches the photon's wave relation E=hf and p=h/λ.
Step 2: Assume the same wave-particle duality for matter
If a massive particle also has a wave associated with it, then its energy should also be E=hf, where f is the frequency of the matter wave. Equating the relativistic energy with the quantum energy:
hf=p2c2+m02c4
For a particle moving at non-relativistic speeds (v≪c), the momentum p=mv is small compared to m0c, so we can expand:
E≈m0c2+2m0p2
The first term is rest energy, which is constant. The second term is kinetic energy K=p2/(2m). The wave frequency f then corresponds to the kinetic part (since rest energy doesn't contribute to motion). But the key relation we want is between wavelength and momentum.
Step 3: The wavelength from the wave speed
For any wave, the speed vwave=fλ. For a matter wave, de Broglie proposed that the wave speed equals the particle's speed v (this is the phase velocity). So:
v=fλ
Now use E=hf and E=21mv2 (non-relativistic kinetic energy). Then:
f=hE=2hmv2
Substitute into v=fλ:
v=2hmv2λ⇒λ=mv2h
This gives λ=2h/p, which is wrong by a factor of 2. The correct formula is λ=h/p. …
Concept: De Broglie Wavelength — every moving particle has a wavelength λ=h/p, where h is Planck’s constant and p is the linear momentum.
Step 1 — Formula
λ=ph=mvh
Step 2 — For the electron
me=9.11×10−31 kg, v=5.4×106 m/s
p=(9.11×10−31)(5.4×106)=4.9194×10−24 kg m/s
λe=4.9194×10−246.626×10−34=1.347×10−10 m
Step 3 — For the ball
m=0.150 kg, v=30.0 m/s …
The de Broglie wavelength is given by λ=h/p. For the electron, λ≈1.35×10−10 m; for the ball, λ≈1.47×10−34 m — the ball’s wavelength is utterly negligible because its mass is huge on the quantum scale.
The core idea here is wave-particle duality. Louis de Broglie proposed that every moving particle has an associated wavelength, just like a photon does. The wavelength is inversely proportional to the particle’s momentum — the heavier or faster the object, the shorter its wavelength. For macroscopic objects like a cricket ball, this wavelength is so tiny that it’s impossible to detect; for electrons, it’s comparable to atomic spacings, which is why electron microscopes work.
The de Broglie wavelength is
λ=ph=mvh
where h=6.63×10−34 J⋅s is Planck’s constant, m is mass in kg, and v is speed in m/s.
Let’s apply this to both cases.
(a) Electron moving at 5.4×106 m/s
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Identify the mass. The electron’s rest mass is me=9.11×10−31 kg. This is a standard value you must remember for such problems.
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Compute momentum.
p=mv=(9.11×10−31)(5.4×106)
Multiply: 9.11×5.4=49.194, and 10−31×106=10−25.
So p=4.9194×10−24 kg⋅m/s.
- Apply de Broglie relation.
λ=ph=4.9194×10−246.63×10−34
Divide: 6.63/4.9194≈1.347, and 10−34/10−24=10−10.
So λ≈1.35×10−10 m.
This wavelength (1.35 A˚) is about the size of an atom. That’s why electron diffraction off crystals is possible — the wavelength matches the spacing between atomic planes.
(b) Ball of mass 150 g at 30.0 m/s
-
Convert mass to kg. 150 g=0.150 kg. This is a common slip — always use SI units.
-
Compute momentum.
p=mv=(0.150)(30.0)=4.50 kg⋅m/s
- Apply de Broglie relation.
λ=4.506.63×10−34
6.63/4.50=1.4733, so
λ≈1.47×10−34 m …
Method: Direct Application of de Broglie's Relation
The de Broglie wavelength λ for any moving particle is given by:
λ=ph=mvh
where h=6.63×10−34 J⋅s is Planck's constant, m is the mass in kg, and v is the speed in m/s.
Steps
Step 1 — Identify the mass and speed for each object.
For (a), the electron: me=9.11×10−31 kg, v=5.4×106 m/s.
For (b), the ball: m=150 g=0.150 kg, v=30.0 m/s.
Step 2 — Write the de Broglie relation and substitute.
For (a):
λ=(9.11×10−31)(5.4×106)6.63×10−34
First compute the denominator:
9.11×10−31×5.4×106=4.9194×10−24 kg⋅m/s.
Then:
λ=4.9194×10−246.63×10−34≈1.35×10−10 m
Step 3 — Repeat for the ball.
For (b):
λ=(0.150)(30.0)6.63×10−34=4.506.63×10−34≈1.47×10−34 m
--- …
Common Mistakes with De Broglie Wavelength Problems
Students typically make the same few errors on this exact type of question. Here's what to watch out for.
1. Forgetting to convert mass to kilograms
The most frequent mistake. Mass must be in kg for the de Broglie formula λ=mvh to work with h=6.63×10−34 J⋅s.
In part (b), the ball's mass is given as 150 g. Many students plug in 150 directly and get a wildly wrong answer.
How to avoid: Always check the units of every quantity before substituting. Convert grams to kilograms by dividing by 1000. Here, 150 g=0.150 kg.
2. Using the wrong value of Planck's constant
Some students use h=6.63×10−34 but forget the exponent, or use h=6.63×1034 (sign error). Others mistakenly use the reduced Planck constant ℏ=2πh.
How to avoid: Memorise h=6.63×10−34 J⋅s as a fixed fact. Write it down before starting the calculation. The de Broglie formula uses h, not ℏ.
3. Mixing up momentum with mass alone
The de Broglie wavelength is λ=ph, where p=mv is momentum. Some students write λ=mh or λ=vh, leaving out velocity or mass.
How to avoid: Write the formula in full every time: λ=mvh. Then substitute both mass and velocity together.
4. Unit errors in the final answer
After calculation, students often report the wavelength in metres when it should be in a more convenient unit like ångströms or nanometres. For an electron, the answer is typically on the order of 10−10 m (1 Å). For a macroscopic ball, it's unimaginably small — around 10−34 m.
How to avoid: Before calculating, estimate the order of magnitude. An electron moving at 106 m/s gives a wavelength near 10−10 m. A 150 g ball at 30 m/s gives 10−34 m. If your answer doesn't match this range, recheck.
5. Forgetting that the electron's mass is known
Some students don't know or forget the electron mass: me=9.1×10−31 kg. They might try to derive it or leave it out.
How to avoid: Memorise me=9.1×10−31 kg as a standard constant for such problems.
Correct Solution
(a) Electron …
Showing the 12 most recent of 16 on this concept.
- KEAM 2026Set eng-2026-04174 marksMCQQ.The kinetic energy of a fast-moving particle of mass 1×10−31 kg is associated with a de Broglie wavelength 63 nm is (h=6.3×10−34 Js) (A) 5×10−21J (B) 1×10−22J (C) 5×10−22J (D) 1×10−21J (E) 2×10−21J
›Reveal solutionSolution
Get momentum from p=h/λ, then kinetic energy from p2/2m: the result is 5×10−22 J.
Momentum:
p=λh=63×10−96.3×10−34=6.3×10−86.3×10−34=1×10−26 kg m s−1.
Kinetic energy: …
- KEAM 2026Set eng-2026-04184 marksMCQQ.For the same kinetic energy, the de Broglie wavelengths associated with particles of different masses are (A) directly proportional to their masses (B) directly proportional to the square root of their masses (C) inversely proportional to the square root of their masses (D) inversely proportional to their masses (E) directly proportional to the square of their masses
›Reveal solutionSolution
The de Broglie wavelength λ=h/2mE, so at fixed kinetic energy it is inversely proportional to m.
de Broglie wavelength in terms of kinetic energy E:
λ=ph=2mEh
For the same kinetic energy E, h and E are constant, so …
- KEAM 2026Set eng-2026-04204 marksMCQQ.If an electron and a proton have same kinetic energy and their de Broglie wavelengths are λe and λp, respectively, then the ratio λp:λe is (A) 1:1 (B) 1:1836 (C) 1836:1 (D) 1836:1 (E) 1:1836
›Reveal solutionSolution
[!TLDR]
At equal kinetic energy the de Broglie wavelength scales as 1/m, so the heavier proton has the shorter wavelength and λp:λe=1:1836.
Concept
The de Broglie wavelength is λ=h/p. For a non-relativistic particle the momentum relates to kinetic energy E by p=2mE, so λ=h/2mE. This wave-nature-of-matter relation is part of the NCERT/CBSE dual-nature chapter.
Solution
For equal kinetic energies E,
λ=2mEh⟹λ∝m1.
Taking the ratio for the proton and electron, …
- KEAM 2026Set eng-2026-04214 marksMCQQ.If the de Broglie wavelength associated with an electron is 0.1227nm, then its accelerating potential is (A) 64V (B) 200V (C) 100V (D) 160V (E) 36V
›Reveal solutionSolution
de Broglie relation for electrons: λ=12.27/V,\u00c5 gives V=100,V.
For an electron accelerated through V volts,
λ=V12.27,0˘0c5. …
- KEAM 2026Set eng-2026-04224 marksMCQQ.If the de Broglie wavelengths of deuteron, positron and electron are in the ratio 1 : 2 : 3, then the ratio of their respective momenta is (A) 1 : 2 : 3 (B) 3:2:1 (C) 6 : 3 : 2 (D) 3 : 2 : 1 (E) 1 : 4 : 9
›Reveal solutionSolution
de Broglie: λ=h/p, so p∝1/λ. Inverting 1:2:3 gives 6:3:2.
From λ=ph, momentum is inversely proportional to wavelength: …
- KEAM 2025Set eng-2025-04234 marksMCQQ.If λ be the wavelength of any electromagnetic radiation, the de-Broglie wavelength of its quantum (photon) is (A) 4λ (B) λ (C) 2λ (D) 2λ (E) 43λ
›Reveal solutionSolution
The de-Broglie wavelength of a photon equals the wavelength lambda of its own radiation.
Concept and Intuition
A photon of electromagnetic wavelength lambda carries momentum p = h/lambda. Its de-Broglie wavelength is defined as lambda_dB = h/p, which returns exactly lambda. So the two coincide for a photon.
Step-by-Step Solution
- Photon momentum p = h/lambda.
- de-Broglie wavelength = h/p = h/(h/lambda) = lambda. …
- KEAM 2025Set eng-2025-04264 marksMCQQ.A sub-atomic particle of mass 6.63×10−31 kg is moving with a velocity of 1×106 ms−1. What is the de Broglie wave length (in nm) associated with it (h = 6.63×10−34 Js)? (A) 10.0 (B) 1.0 (C) 0.10 (D) 5.0 (E) 0.50
›Reveal solutionSolution
The de Broglie wavelength λ=h/(mv) works out to 10−9 m, i.e. 1.0 nm.
The de Broglie relation gives
λ=mvh=(6.63×10−31)(1×106)6.63×10−34 …
- KEAM 2025Set eng-2025-04274 marksMCQQ.The ratio of the respective de Broglie wavelengths of two particles with kinetic energy of 0.02 eV and 2 eV, respectively, is (A) 1:1 (B) 10:1 (C) 1:10 (D) 1:10 (E) 10:1
›Reveal solutionSolution
Since λ∝K1 for equal masses, the wavelength ratio is K2/K1=2/0.02=10:1.
The de Broglie wavelength in terms of kinetic energy K is:
λ=ph=2mKh.
For particles of equal mass, λ∝K1. Taking K1=0.02 eV and K2=2 eV: …
- KEAM 2025Set pha-2025-0424F4 marksMCQQ.A sub-atomic particle of mass 2.2×10−2 kg is moving with a velocity of 3.0×105 ms−1. What is its de Broglie wavelength? (Planck's constant h = 6.6×10−34 Js) (A) 1 pm (B) 0.1 pm (C) 2 pm (D) 0.2 pm (E) 0.5 pm
›Reveal solutionSolution
[!TLDR]
Applying de Broglie's relation λ=h/mv, the numerical coefficients divide to exactly 1 and the scale works out to picometres, giving λ=1 pm.
Concept
Louis de Broglie proposed that every moving particle has an associated wavelength λ=mvh, where h is Planck's constant and mv is the momentum. This wave-particle duality is a core idea in the NCERT/CBSE-aligned 'Structure of Atom' chemistry syllabus that KEAM follows.
Solution
λ=mvh=(2.2×10−27)(3.0×105)6.6×10−34.
The coefficients divide cleanly:
2.2×3.06.6=6.66.6=1.
The powers of ten give
10−27⋅10510−34=10−34+27−5=10−12 m.
Hence
λ=1×10−12 m=1 pm. …
- KEAM 2025Set pha-2025-0429F4 marksMCQQ.If a particle is moving with a momentum of (2×1010)h kgms−1 then the de Broglie wavelength associated with it (in angstrom) is (where h is Planck’s constant) (A) 1.5 (B) 2.5 (C) 1.0 (D) 0.5 (E) 0.75
›Reveal solutionSolution
The de Broglie wavelength λ=h/p; here p=(2×1010)h, so λ=1/(2×1010)m=0.5A˚.
The de Broglie wavelength is
λ=ph.
Given p=(2×1010)h kgms−1, the Planck constant cancels: …
- KEAM 2024Set eng-2024-06054 marksMCQQ.The de Broglie wavelength associated with the electrons accelerated by a potential of 81 V is lying in the region of electromagnetic waves (A) ultraviolet rays (B) infrared rays (C) microwaves (D) X-rays (E) γ-rays
›Reveal solutionSolution
λ≈1.36 Å for 81 V electrons ⇒ X-ray region.
The de Broglie wavelength of an electron accelerated through potential V is
λ=V1.227 nm=811.227=91.227=0.136 nm=1.36 A˚. …
- KEAM 2024Set eng-2024-06064 marksMCQQ.A particle having mass 2000 times that of an electron travels with a velocity thrice that of the electron. The ratio of the de Broglie wavelength of the particle to that of the electron is (A) 30001 (B) 20001 (C) 60001 (D) 80001 (E) 15001
›Reveal solutionSolution
λ=h/mv, so the ratio is 1/(2000×3)=1/6000.
The de Broglie wavelength is λ=mvh, inversely proportional to the product mv. …
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